MIDTERM 1 FOR Multivariable Calculus, Math 251(22-24), FALL 2020, Dr. Z. NAME:Liuyang Shan RUID:203002003 EMAIL:ls1225@rutgers.edu BELOW WRITE THE LIST OF THE ANSWERS Answer[ 1 ]=dz/dy=-2/5 Answer[ 2 ]=f is increasing at the direction <2,3,-1> (WRONG ANS. RIGHT WAY, -5 POINTS) Answer[ 3 ]=directional derivative=3 Answer[ 4 ]=(1,2) Answer[ 5 ]=4 Answer[ 6 ]=doesn't exist (WRONG ANS., WRONG WAY -10 POINTS) Answer[ 7 ]=6 Answer[ 8 ]=p(pi)=[0,-1] Answer[ 9 ]=5 Answer[ 10 ]= (1,1,3) SCORE: 85 POINTS) ----------------------------------------------------------------- Instructions: Download this file with its original name, mt1.txt, then rename it, in your computer mt1FirstLast.txt Edit it with your answers and solutions USING COMPUTEREZE: e.g.: x times y IS x*y, x to the power y is x^y and Email DrZcalc3@gmail.com, 80 minutes (or sooner) after starting (for most people 10:00am, Oct. 15) Subject: mt1 with an attachment. YOU MUST NAME IT EXACTLY mt1FirstLast.txt --------------------------------------------------------------------------- For each of the questions you MUST first figure, YOUR version, with the following convention For i=1,2,3,4,5,6,7,8,9 , a[i]:= The i-th digit of your RUID, BUT of it is zero make it 1 Example: RUID=413200125; a[1] = 4, a[2] = 1, a[3] = 3, a[4] = 2, a[5] = 1, a[6] = 1, a[7] = 1, a[8] = 2, a[9] = 5 ----------------------------------------------------------------------------------------------------------------------------------------------- HERE WRITE THE ACTUAL a[i] a[1]=2 , a[2]=1 , a[3]=3 , a[4]=1 , a[5]=1 , a[6]=2 , a[7]=1 , a[8]=1 , a[9]=3 -------------------------------------------- --------------------------------------------- Problem 1: Find dz/dy at the point (1,1,1) if z(x,y) is given implicitly by the equation x^a[1]+y^a[2]+z^a[3]+a[5]*x*y*z^2 = 3+a[5] With my RUID data the question is x^2+y+z^3+x*y*z^2=4 Here is how I do it (Explain everything) 0+1+3*z^2*dz/dy+z^2+2*z*y*dz/dy=0 and plugging (1,1,1) into this equation 2+5*dz/dy=0 Ans.: dz/dy=-2/5 --------------------------------------------- Problem 2: Suppose that grad(f)(P)=. Is f increasing or decreasing at the direction ? With my RUID data the question is grad(f)(P)=<2,-1,3> direction <2,3,-1> Here is how I do it (Explain everything) the dot product of grad(f)(P) and direction vector is 2*2+(-1)*3+3*(-1)=4>0 Ans.: f is increasing at the direction <2,3,-1> --------------------------------------------- Problem 3: Find the directional derivative of the function f(x,y,z) x^3*a[6]+y^3*a[3]+z^3*a[8] At the point P=(1,-1,1) in the direction pointing to Q=(1,-1,3) With my RUID data the question is f(x,y,z)=2*x^3+3*y^3+z^3 Here is how I do it (Explain everything) grad f(x,y,z)=(6*x^2, 9*y^2, 3*z^2) grad f(1,-1,1)=(6, 9, 3) direction vector is u=(0,0,1) the directional derivative is dot product of del f(6, 9, 3) and u=(0,0,1), which equals to 3 Ans.: 3 --------------------------------------------- Problem 4: Find a saddle point of the function f(x,y)= exp(x-a[4])-(x-a[4])*exp(y-a[6]) If there is no saddle point, write in the Answers: "Does Not exist". Explain what you are doing With my RUID data the question is f(x,y)=exp(x-1)-(x-1)*exp(y-2) Here is how I do it (Explain everything) f_x(x,y)=exp(x-1)-exp(y-2) f_y(x,y)=-(x-1)*exp(y-2), so the critical point is (1,2) f_xx(x,y)=exp(x-1) f_yy(x,y)=-(x-1)*exp(y-2) f_xy(x,y)=-exp(y-2) D=f_xx(x,y)*f_yy(x,y)-(f_xy(x,y)=-exp(y-2))^2=0-(-1)^2=-1. D<0, so (1,2) is the saddle point. Ans.: (1,2) --------------------------------------------- Problem 5: Let f(x,y) be the function a[4]*x + a[7]*y + a[2] Find the ABSOLUTE MINIMUM VALUE of f(x,y) INSIDE the TRIANGLE whose VERTICES ARE A = [a[1], a[2]], B = [a[3], a[4]], C = [a[5], a[6]] With my RUID data the question is f(x,y)=x+y+1 A=[2,1] B=[3,1] C=[1,2] Here is how I do it (Explain everything) L_AC:x+y=3 (x=1..2) L_AB: y=1 (x=2..3) L_BC:x+2y=5 (x=1..3). Because AC and AB is the left and bottom margin of triangle ABC, for each point (a,b) on the AC(or AB) , a(or b) is the minimum value of x(or y) on the line y=b(or x=a) inside the triangle. So, the point that has minimun of value of x+y inside the triangle must on the AB or AC. Apparantly, the minimum of x+y is 3. So, the minimun value of f(x,y)=x+y+1 inside the triangle ABC is 4. Ans.: 4 --------------------------------------------- Problem 6: Let f(x,y) be the function (x^2*a[4]^2-y^2*a[5]^2)/(x*a[4]-y*a[5]) Find the LIMIT of f(x,y) as (x,y) goes to the point [a[5],a[4]], or show that it does not exist With my RUID data the question is f(x,y)=2*(x^2-y^2)/(x-y), the point [1,1] Here is how I do it (Explain everything) Plugging y=k*x into the equation, f(x,y)=2*(1-k^2)*x^2/((1-k)*x)=2*(1+k)*x whose value is related to k. So, the LIMIT of f(x,y) goes to the point [1,1] doesn't exist. --------------------------------------------- Problem 7: Find the curvature of the curve r(t) = [a[1], a[2]*t, a[3]*t^2] At the point (a[1],0,0) With my RUID data the question is r(t)=[2,t,3*t^2] the point (2,0,0) Here is how I do it (Explain everything) r'(t)=[0,1,6*t]=[0,1,0] r''(t)=[0,0,6] k=|r'(t)×r''(t)|/(|r'(t)|)^3=6/1=6 --------------------------------------------- Problem 8: A particle is moving in the plane with ACCELERATION given by [-a[1]*sin(t), -a[2]*cos(t)] At time t=0 its position is , [0, a[2]] and its velocity is , [a[1], 0] Where is it located at time , t = Pi With my RUID data the question is a(t)=[-2*sin(t), -cos(t)] p(0)=[0,1], v(0)=[2,0] Here is how I do it (Explain everything) v(t)=[2*cos(t), -sin(t)] p(t)=[2*sin(t), cos(t)] p(pi)=[0,-1] --------------------------------------------- Problem 9: A certain function depends on variables x and y Right now the rate of change of the function with respect to x is, a[5] and the rate of change of the function with respect to y is, a[7] Both x and y depend on time Right now the rate of change of x with respect to time is, a[1] and the rate of change of y with respect to time is, a[9] How fast is the function changing right now? With my RUID data the question is f_x(x,y)=1 f_y(x,y)=1 dx/dt=2 dy/dt=3 Here is how I do it (Explain everything) df/dt=(df/dx)*(dx/dt)+(df/dy)*(dy/dt)=2+3=5 --------------------------------------------- Problem 10: Find the point of intersection of the three planes x = a[5], y = a[7], z = a[3] With my RUID data the question is x=1 y=1 z=3 Here is how I do it (Explain everything) point (1,1,3)